Question

What is the area with respect to the y axis of the curve x² = 3y, from x = 0 to x = 4?

Answer 1

Let us find the area $A$ with respect to the y-axis of the curve $x^2 = 3y,$ from $x = 0$ to $x = 4.$ If $x=0$then $y=0.$ If $x=4$ then $y=\frac{16}3.$ It follows that $x=\pm\sqrt{3y},$ and using the symmetry of the curve about the y-axis, we get that

$A=2\int\limits_0^{\frac{16}3}\sqrt{3y}dy =2\sqrt{3}\frac{2}3y^{\frac{3}2}|_0^{\frac{16}3} =\frac{4}3\sqrt{3}(\frac{16}3)^{\frac{3}2} =\frac{4}3\sqrt{3}\frac{4^3}{3\sqrt{3}} =\frac{256}{9}$ (sq. units).